What is the inverse of the function $h(x)=\dfrac{3-x}{x+1}$ ? $ h^{-1}(x) =$
Let's start by replacing $h(x)$ with $y$. $y=\dfrac{3-x}{x+1}$ Now let's swap $x$ and $y$ and solve for $y$. $\dfrac{3-y}{y+1}=x$ [Why do we swap x and y?] $\begin{aligned} \dfrac{3-y}{y+1}&=x \\\\ 3-y&=x(y+1) \\\\ 3-y&=xy+x \\\\ 3-x&=xy+y \\\\ 3-x&=y(x+1) \\\\ y&=\dfrac{3-x}{x+1} \end{aligned}$ In conclusion, this is the inverse function: $h^{-1}(x)=\dfrac{3-x}{x+1}$ Note: This is an example for which a function is its own inverse! [I saw someone solve this problem by originally solving for x. Were they wrong?]